Hermite polynomials from binomial theorem Hello I want to obtain the explicit formula for the Hermite polynomial:
$$
H_n(x)=\sum_{k=0}^{[n/2]}\frac{(-1)^k n!}{k! (n-2k)!} (2x)^{n-2k}
$$
by expanding the exponential in the generating function
$$
e^{2xz-z^2}=\sum_{n=0}^\infty H_{n}(x)\frac{z^n}{n!}
$$
and using the binomial theorem
$$
e^{2xz-z^2}=\sum_{n=0}^\infty \frac{(2xz-z^2)^n}{n!}=\sum_{n=0}^\infty \sum_{k=0}^{n} \frac{(-1)^k n! }{k!(n-k)!} \frac{(2xz)^{n-k}z^{2k}}{n!}=\sum_{n=0}^\infty \sum_{k=0}^{n} \frac{(-1)^kn!}{k!(n-k)!} (2x)^{n-k} \frac{z^{n+k}}{n!}
$$
It seems almost done, but what is the easiest way to transform the last result in the Hermite polynomial?
 A: The last step is differentiation of boths sides with respect to $z$, as
$$
\frac{\partial^m}{\partial z^m}e^{2xz-z^2}|_{z=0}\equiv H_m(x),
$$
by the definition of the generating function.
Performing the differentiation leads to
$$
H_m(x) = \sum_{n=0}^\infty \sum_{k=0}^n \frac{(-1)^k}{k!(n-k)!}(2x)^{n-k}(n+k)(n+k-1)\cdots(n+k-m+1)z^{n+k-m}|_{z=0}.
$$
The last expression is only non-zero when $n+k=m$, restricting ones of the sums. This also restricts the parity of the polynomial, as $\operatorname{mod}(n,2)+\operatorname{mod}(k,2) = \operatorname{mod}(m,2)$, which also implies $\operatorname{mod}(n,2)-\operatorname{mod}(n,2)=\operatorname{mod}(n,2)$ meaning the parity of the polynomial inside the sum is the same as the parity of $m$, as required.
Note that $n+k=m$ but $n\ge k$ (since the upper limit of summation over $k$ is $n$) therefore the the minimum value $k$ can take is $0$ when $n=m$ and the maximum is $\left \lfloor \frac{m}{2} \right \rfloor $ when $n=k$.  This can be seen graphically if you plot out the $(k,n)$ coordinates:

Making this substitution leads to the formula, since the term $(n+k)(n+k-1)\cdots(n+k-m+1) = \frac{(n+k)!}{(n+k-m)!}$ becomes simply $m!$ by the substitution $n=m-k$, and resulting expression is
$$
H_m(x)=m!\sum_{k=0}^{\left \lfloor \frac{m}{2} \right \rfloor} \frac{(-1)^k}{k!(n-2k)!}(2x)^{n-2k}
$$
as desired.  Note your original equation is missing the leading $n!$.  I switched to $m$ because you already had $n$ as a summation index.
